No, we need some other information like if this quadrilateral is part of any circle or some other way in which we can differentiate x and y But if we use only these three equal sides then we only get x + y = 72°(in all cases)
In isosceles triangle ABC,
∠B = ∠C = \(\frac{180^{\circ}-24^{\circ}}2\) = \(\frac{156^{\circ}}2=78^{\circ}\)
∴ ∠5 = 78° - 48° = 30°
∠6 = ∠C = 78°
In Δ BCD, ∠5 + ∠6 + y + x = 180°
⇒ x + y = 180° - 30° - 78° = 72°
In isosceles triangle ABD
∠A + ∠4 = ∠3 = \(\frac{180^{\circ}-48^{\circ}}2=\frac{132^{\circ}}2=60^{\circ}\)
∴ ∠3 = 66°
∠1 = 66° - ∠A = 66° - 24° = 42°
with its help we also can get (in Δ ACD)
x + y = 72°
But not a separate value of x & y.
∵ x + y = 72° and x < y (∵ CE < DE)
∴ x < 36°
(If options are given then can apply it otherwise did not get the logic which solve it )